Question
Let $X_1, \ldots, X_n$ be a random sample from an exponential distribution, $X \sim \operatorname{EXP}(\theta)$ If $\bar{x}=17.9$ with $n=50$, then find a one-sided lower $95 \%$ confidence limit for $\theta$.
Solution
The pivotal quantity $\frac{2 n \bar{X}}{\theta} \sim \chi^2(2 n)$ yields $\mathbb{P}\left(\frac{2 n \bar{X}}{\theta}<\chi_\gamma^2\right)=\gamma$ and $\mathbb{P}\left(\frac{2 n \bar{X}}{\chi_\gamma^2}<\theta\right)=\gamma$. With $\chi_\gamma^2(2 n)=\chi_{0.95}^2(100)=124.34$, a one-sided lower $95 \%$ confidence limit for $\theta$ is obtained as $$ \ell\left(x_1, \ldots, x_n\right)=\frac{2 n \bar{x}}{\chi_\gamma^2}=\frac{2 \cdot 50 \cdot 17.9}{124.34}=14.396 $$
Problem
Currently I am learning about pivotal quantities and the fact that a pivotal quantity is only a function of the data points and the unknown parameter (in this case $\theta$). What I don't understand yet is how the expression $\frac{2 n \bar{X}}{\theta} \sim \chi^2(2 n)$ is derived from the data given in the exercise.
Questions:
- How is the pivotal quantity derived?
- Why is it Chi-squared distributed?